Question: $ \lim_{x\to 5}\dfrac{3-x^2}{2x+1}=$
Explanation: $\dfrac{3-x^2}{2x+1}$ defines a rational function. Rational functions are continuous across their entire domain, and their domain is all real $x$ -values that don't make the denominator equal to zero. In other words, for any rational function $r$ and any input $c$ in the domain of $r$, we know that this equality holds: $\lim_{x\to c}r(x)=r(c)$ The input $x=5$ is within the domain of $\dfrac{3-x^2}{2x+1}$. Therefore, in order to find $ \lim_{x\to 5}\dfrac{3-x^2}{2x+1}$, we can simply evaluate $\dfrac{3-x^2}{2x+1}$ at $x=5$. $\begin{aligned} &\phantom{=}\dfrac{3-x^2}{2x+1} \\\\ &=\dfrac{3-(5)^2}{2(5)+1} \gray{\text{Substitute }x=5} \\\\ &=\dfrac{-22}{11} \\\\ &=-2 \end{aligned}$ In conclusion, $ \lim_{x\to 5}\dfrac{3-x^2}{2x+1}=-2$.